PROTUGUESE_BANK_PROJECT_1¶

DOMAIN ANALYSIS¶

INPUT VARIABLES:¶

ATTRIBUTES INFORMATION:¶

1 - age (numeric)¶

2 - job : type of job (categorical: 'admin.','blue-collar','entrepreneur','housemaid','management','retired','self-employed','services','student','technician','unemployed','unknown')¶

3 - marital : marital status (categorical: 'divorced','married','single','unknown'; note:'divorced' means divorced or widowed)¶

4 - education (categorical:'basic.4y','basic.6y','basic.9y','high.school','illiterate','professional.course','university.degree','unknown')¶

5 - default: has credit in default? (categorical: 'no','yes','unknown')¶

6 - housing: has housing loan? (categorical: 'no','yes','unknown')¶

7 - loan: has personal loan? (categorical: 'no','yes','unknown')¶

Related with the last contact of the current campaign:¶

8 - contact: contact communication type (categorical: 'cellular','telephone')¶

9 - month: last contact month of year (categorical: 'jan', 'feb', 'mar', ..., 'nov', 'dec')¶

10 - day_of_week: last contact day of the week (categorical: 'mon','tue','wed','thu','fri')¶

11 - duration: last contact duration, in seconds (numeric). Important note: this attributehighly affects the output target (e.g., if duration=0 then y='no'). Yet, the duration is notknown before a call is performed. Also, after the end of the call y is obviously known.Thus, this input should only be included for benchmark purposes and should bediscarded if the intention is to have a realistic predictive model.¶

Other Attributes:¶

12 - campaign: number of contacts performed during this campaign and for this client(numeric, includes last contact)¶

13 - pdays: number of days that passed by after the client was last contacted from aprevious campaign (numeric; 999 means client was not previously contacted)¶

14 - previous: number of contacts performed before this campaign and for this client(numeric)¶

15 - poutcome: outcome of the previous marketing campaign (categorical:'failure','nonexistent','success')¶

Social and economic context attributes:¶

16 - emp.var.rate: employment variation rate - quarterly indicator (numeric)¶

17 - cons.price.idx: consumer price index - monthly indicator (numeric)¶

18 - cons.conf.idx: consumer confidence index - monthly indicator (numeric)¶

19 - euribor3m: euribor 3 month rate - daily indicator (numeric)¶

20 - nr.employed: number of employees - quarterly indicator (numeric)¶

OUTPUT VARIABLES(Desired target):¶

21 - y - has the client subscribed a term deposit? (binary: 'yes','no')¶

In [1]:
# importing libraries

import pandas as pd
import numpy as np
import seaborn as sns 
import matplotlib.pyplot as plt
In [2]:
data = pd.read_csv('bank-additional-full.csv', sep=';')

BASICS CHECK¶

In [3]:
data.head() # it give top 5 data
Out[3]:
age job marital education default housing loan contact month day_of_week ... campaign pdays previous poutcome emp.var.rate cons.price.idx cons.conf.idx euribor3m nr.employed y
0 56 housemaid married basic.4y no no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 no
1 57 services married high.school unknown no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 no
2 37 services married high.school no yes no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 no
3 40 admin. married basic.6y no no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 no
4 56 services married high.school no no yes telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 no

5 rows × 21 columns

In [4]:
data.tail() # it's give last 5 rows
Out[4]:
age job marital education default housing loan contact month day_of_week ... campaign pdays previous poutcome emp.var.rate cons.price.idx cons.conf.idx euribor3m nr.employed y
41183 73 retired married professional.course no yes no cellular nov fri ... 1 999 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 yes
41184 46 blue-collar married professional.course no no no cellular nov fri ... 1 999 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 no
41185 56 retired married university.degree no yes no cellular nov fri ... 2 999 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 no
41186 44 technician married professional.course no no no cellular nov fri ... 1 999 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 yes
41187 74 retired married professional.course no yes no cellular nov fri ... 3 999 1 failure -1.1 94.767 -50.8 1.028 4963.6 no

5 rows × 21 columns

In [5]:
data.columns
Out[5]:
Index(['age', 'job', 'marital', 'education', 'default', 'housing', 'loan',
       'contact', 'month', 'day_of_week', 'duration', 'campaign', 'pdays',
       'previous', 'poutcome', 'emp.var.rate', 'cons.price.idx',
       'cons.conf.idx', 'euribor3m', 'nr.employed', 'y'],
      dtype='object')

Replacing the categorical data of y to numerical¶

In [6]:
data['y'] = data['y'].replace({'yes':1, 'no':0})  # here replacing categorical to numerical
data['y'] = data['y'].astype('int64')
data
Out[6]:
age job marital education default housing loan contact month day_of_week ... campaign pdays previous poutcome emp.var.rate cons.price.idx cons.conf.idx euribor3m nr.employed y
0 56 housemaid married basic.4y no no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
1 57 services married high.school unknown no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
2 37 services married high.school no yes no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
3 40 admin. married basic.6y no no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
4 56 services married high.school no no yes telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
41183 73 retired married professional.course no yes no cellular nov fri ... 1 999 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 1
41184 46 blue-collar married professional.course no no no cellular nov fri ... 1 999 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 0
41185 56 retired married university.degree no yes no cellular nov fri ... 2 999 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 0
41186 44 technician married professional.course no no no cellular nov fri ... 1 999 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 1
41187 74 retired married professional.course no yes no cellular nov fri ... 3 999 1 failure -1.1 94.767 -50.8 1.028 4963.6 0

41188 rows × 21 columns

I will also rename some columns replacing a dot by underscores¶

In [7]:
data.rename(columns= {'emp.var.rate':'emp_var_rate', 'cons.price.idx': 'cons_price_idx', 'cons.conf.idx':'cons_conf_idx',
                      'nr.employed':'nr_employed'}, inplace = True)
In [8]:
data.head()
Out[8]:
age job marital education default housing loan contact month day_of_week ... campaign pdays previous poutcome emp_var_rate cons_price_idx cons_conf_idx euribor3m nr_employed y
0 56 housemaid married basic.4y no no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
1 57 services married high.school unknown no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
2 37 services married high.school no yes no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
3 40 admin. married basic.6y no no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
4 56 services married high.school no no yes telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0

5 rows × 21 columns

In [9]:
data.describe()  # it's give numerical data here, we find std, mean, median, min and max values
Out[9]:
age duration campaign pdays previous emp_var_rate cons_price_idx cons_conf_idx euribor3m nr_employed y
count 41188.00000 41188.000000 41188.000000 41188.000000 41188.000000 41188.000000 41188.000000 41188.000000 41188.000000 41188.000000 41188.000000
mean 40.02406 258.285010 2.567593 962.475454 0.172963 0.081886 93.575664 -40.502600 3.621291 5167.035911 0.112654
std 10.42125 259.279249 2.770014 186.910907 0.494901 1.570960 0.578840 4.628198 1.734447 72.251528 0.316173
min 17.00000 0.000000 1.000000 0.000000 0.000000 -3.400000 92.201000 -50.800000 0.634000 4963.600000 0.000000
25% 32.00000 102.000000 1.000000 999.000000 0.000000 -1.800000 93.075000 -42.700000 1.344000 5099.100000 0.000000
50% 38.00000 180.000000 2.000000 999.000000 0.000000 1.100000 93.749000 -41.800000 4.857000 5191.000000 0.000000
75% 47.00000 319.000000 3.000000 999.000000 0.000000 1.400000 93.994000 -36.400000 4.961000 5228.100000 0.000000
max 98.00000 4918.000000 56.000000 999.000000 7.000000 1.400000 94.767000 -26.900000 5.045000 5228.100000 1.000000
In [ ]:
In [10]:
data.describe(include='O') # here, we describe categorical data
Out[10]:
job marital education default housing loan contact month day_of_week poutcome
count 41188 41188 41188 41188 41188 41188 41188 41188 41188 41188
unique 12 4 8 3 3 3 2 10 5 3
top admin. married university.degree no yes no cellular may thu nonexistent
freq 10422 24928 12168 32588 21576 33950 26144 13769 8623 35563
In [11]:
data.info(), # Describing datatypes, entries and non-null values

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 41188 entries, 0 to 41187
Data columns (total 21 columns):
 #   Column          Non-Null Count  Dtype  
---  ------          --------------  -----  
 0   age             41188 non-null  int64  
 1   job             41188 non-null  object 
 2   marital         41188 non-null  object 
 3   education       41188 non-null  object 
 4   default         41188 non-null  object 
 5   housing         41188 non-null  object 
 6   loan            41188 non-null  object 
 7   contact         41188 non-null  object 
 8   month           41188 non-null  object 
 9   day_of_week     41188 non-null  object 
 10  duration        41188 non-null  int64  
 11  campaign        41188 non-null  int64  
 12  pdays           41188 non-null  int64  
 13  previous        41188 non-null  int64  
 14  poutcome        41188 non-null  object 
 15  emp_var_rate    41188 non-null  float64
 16  cons_price_idx  41188 non-null  float64
 17  cons_conf_idx   41188 non-null  float64
 18  euribor3m       41188 non-null  float64
 19  nr_employed     41188 non-null  float64
 20  y               41188 non-null  int64  
dtypes: float64(5), int64(6), object(10)
memory usage: 6.6+ MB
Out[11]:
(None,)

here, no null values¶

In [12]:
data.dtypes
Out[12]:
age                 int64
job                object
marital            object
education          object
default            object
housing            object
loan               object
contact            object
month              object
day_of_week        object
duration            int64
campaign            int64
pdays               int64
previous            int64
poutcome           object
emp_var_rate      float64
cons_price_idx    float64
cons_conf_idx     float64
euribor3m         float64
nr_employed       float64
y                   int64
dtype: object
In [13]:
data.shape #  describing how many rows and columns. here, 41188 are rows and 21 are columns
Out[13]:
(41188, 21)
In [14]:
data.duplicated().sum() # describing duplicate
Out[14]:
12

I must conclude that these apparent duplicated samples are actually from people with an identical profile.¶

In [15]:
data.y.value_counts()[0] # Number of client that haven't subscribed the term deposit
Out[15]:
36548
In [16]:
data.y.value_counts()[1] # Number of clients that have subscribed the term deposit
Out[16]:
4640

[0] Number of clients that haven’t subscribed the term deposit: 36548¶

[1] Number of clients that have subscribed the term deposit: 4640¶

*After basics check, we do exploratory data analysis to see didtribution of data through different graph¶

EXPLORATORY DATA ANALYSIS¶

Univariate Analysis¶

In [17]:
data.head()
Out[17]:
age job marital education default housing loan contact month day_of_week ... campaign pdays previous poutcome emp_var_rate cons_price_idx cons_conf_idx euribor3m nr_employed y
0 56 housemaid married basic.4y no no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
1 57 services married high.school unknown no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
2 37 services married high.school no yes no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
3 40 admin. married basic.6y no no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
4 56 services married high.school no no yes telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0

5 rows × 21 columns

In [18]:
# Analyzing a age through univariate
plt.figure(figsize=(20,20))
sns.countplot(x=data['age'], data=data)
plt.show()
In [19]:
data.columns
Out[19]:
Index(['age', 'job', 'marital', 'education', 'default', 'housing', 'loan',
       'contact', 'month', 'day_of_week', 'duration', 'campaign', 'pdays',
       'previous', 'poutcome', 'emp_var_rate', 'cons_price_idx',
       'cons_conf_idx', 'euribor3m', 'nr_employed', 'y'],
      dtype='object')
In [20]:
# Analyzing the all numerical columns using histplot

import warnings
warnings.filterwarnings('ignore')
data1=data[['age', 'duration', 'campaign', 'pdays', 'previous', 'emp_var_rate','cons_price_idx', 'cons_conf_idx', 'euribor3m', 'nr_employed','y']]
plt.figure(figsize=(10,100),facecolor="white")
plotnumber=1
for column in data1:
    
    if plotnumber<=10:
        
        ax=plt.subplot(30,1,plotnumber)
        sns.countplot(x=data1[column])
        plt.xlabel(column,fontsize=20)
        
        
    plotnumber+=1
plt.tight_layout()
plt.show()
In [21]:
# analyze the categorical column using histplot 
import warnings
warnings.filterwarnings('ignore')
data2=data[['job', 'marital', 'education', 'default', 'housing', 'loan', 'contact', 'month', 'day_of_week', 'poutcome']]
plt.figure(figsize=(10,100), facecolor= 'white')
plotnumber=1

for column in data2:
    if plotnumber<=11:
        ax=plt.subplot(20,1, plotnumber)
        sns.countplot(x=data2[column])
        plt.xlabel(column, fontsize=20)
        
    plotnumber+=1
    
plt.tight_layout()
plt.show()
In [22]:
# Analyzing the numerical data using pieolot
data1 = data[['age', 'duration', 'campaign', 'pdays', 'previous', 'emp_var_rate','cons_price_idx', 'cons_conf_idx', 'euribor3m', 'nr_employed','y']]
plt.figure(figsize=(200,200))
plotnumber=1

for column in data1:
    if plotnumber <= 10:
        ax = plt.subplot(21,1, plotnumber)
        ax.pie(data1[column].value_counts(normalize=True), labels=data[column].value_counts().index,autopct='%1.1f%%')
        ax.set_title(column,fontsize=25)
    plotnumber+=1
plt.tight_layout()
plt.show()
In [23]:
# Analyzing categorical data using pieplot
data2=data[['job', 'marital', 'education', 'default', 'housing', 'loan', 'contact', 'month', 'day_of_week', 'poutcome']]
plt.figure(figsize=(100,100))
plotnumber=1

for column in data2:
    if plotnumber<=10:
        ax = plt.subplot(21,1, plotnumber)
        ax.pie(data2[column].value_counts(normalize=True), labels=data[column].value_counts().index, autopct='%1.1f%%')
        ax.set_title(column, fontsize=20)
    plotnumber+=1
plt.tight_layout()
plt.show()
In [24]:
# Analysing the numerical data using histplot
import warnings
warnings.filterwarnings('ignore')
data1=data[['age', 'duration', 'campaign', 'pdays', 'previous', 'emp_var_rate','cons_price_idx', 'cons_conf_idx', 'euribor3m', 'nr_employed','y']]
plt.figure(figsize=(10,80))
plotnumber=1

for column in data1:
    if plotnumber <=10:
        ax=plt.subplot(10,1, plotnumber)
        sns.histplot(x=data1[column], hue=data.y, kde=True)
        plt.xlabel(column, fontsize=25)
        plt.ylabel('y', fontsize=25)
    plotnumber+=1
plt.show()
In [25]:
# Analyzing categorical data using histplot
data2=data[['job', 'marital', 'education', 'default', 'housing', 'loan', 'contact', 'month', 'day_of_week', 'poutcome']]
plt.figure(figsize=(20,200))
plotnumber=1

for column in data2:
    if plotnumber<=10:
        ax=plt.subplot(10,1,plotnumber)
        sns.histplot(x=data2[column], kde=True, hue=data.y)
        plt.xlabel(column, fontsize=20)
        plt.ylabel('y', fontsize=20)
    plotnumber+=1
plt.show()
In [26]:
# Analyzing Numerical analysis using distplot
plt.figure(figsize=(10,100), facecolor='white')
data1=data[['age', 'duration', 'campaign', 'pdays', 'previous', 'emp_var_rate','cons_price_idx', 'cons_conf_idx', 'euribor3m', 'nr_employed','y']]
plotnumber=1

for column in data1:
    if plotnumber<=10:
        ax=plt.subplot(10,1, plotnumber)
        sns.distplot(x=data1[column])
        plt.xlabel(column, fontsize=20)
    plotnumber+=1
plt.show()
In [27]:
# Checking outliers using boxplot
import warnings
warnings.filterwarnings('ignore')
data1=data[['age', 'duration', 'campaign', 'pdays', 'previous', 'emp_var_rate','cons_price_idx', 'cons_conf_idx', 'euribor3m', 'nr_employed']]
plt.figure(figsize=(10,150), facecolor='white')
plotnumber=1

for column in data1:
    if plotnumber<=11:
        ax=plt.subplot(11,1, plotnumber)
        sns.boxplot(data1[column])
        plt.xlabel(column, fontsize=20)
    plotnumber+=1
plt.show()
In [28]:
!pip install sweetviz

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In [29]:
import sweetviz as sv
my_report=sv.analyze(data)
my_report.show_html()

                                             |          | [  0%]   00:00 -> (? left)
Report SWEETVIZ_REPORT.html was generated! NOTEBOOK/COLAB USERS: the web browser MAY not pop up, regardless, the report IS saved in your notebook/colab files.

BIVARIATE ANALYSIS¶

In [30]:
data
Out[30]:
age job marital education default housing loan contact month day_of_week ... campaign pdays previous poutcome emp_var_rate cons_price_idx cons_conf_idx euribor3m nr_employed y
0 56 housemaid married basic.4y no no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
1 57 services married high.school unknown no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
2 37 services married high.school no yes no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
3 40 admin. married basic.6y no no no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
4 56 services married high.school no no yes telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
41183 73 retired married professional.course no yes no cellular nov fri ... 1 999 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 1
41184 46 blue-collar married professional.course no no no cellular nov fri ... 1 999 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 0
41185 56 retired married university.degree no yes no cellular nov fri ... 2 999 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 0
41186 44 technician married professional.course no no no cellular nov fri ... 1 999 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 1
41187 74 retired married professional.course no yes no cellular nov fri ... 3 999 1 failure -1.1 94.767 -50.8 1.028 4963.6 0

41188 rows × 21 columns

In [31]:
# Bivariate analyzing using countplot
import warnings
warnings.filterwarnings('ignore')
plt.figure(figsize=(20,50), facecolor='white')
data1=data[['age', 'duration', 'campaign', 'pdays', 'previous', 'emp_var_rate','cons_price_idx', 'cons_conf_idx', 'euribor3m', 'nr_employed']]
plotnumber=1

for column in data1:
    if plotnumber<=11:
        ax=plt.subplot(10,2, plotnumber)
        sns.countplot(x=data1[column], hue=data['y'])
        plt.xlabel(column, fontsize=20)
    plotnumber+=1
plt.show()

MULTIVARIATE ANALYSIS¶

In [32]:
sns.pairplot(data,hue='y',palette='crest', size=3)
Out[32]:
<seaborn.axisgrid.PairGrid at 0x1a5df828850>

INSIGHTS IN THE EDA(Exploratory Data Analysis¶

Types of data¶

Variable types:-¶

Numeric:-['age', 'duration', 'campaign', 'pdays', 'previous', 'emp_var_rate','cons_price_idx', 'cons_conf_idx', 'euribor3m', 'nr_employed' 'y']¶

Categorical:-['job', 'marital', 'education', 'default', 'housing', 'loan', 'contact', 'month', 'day_of_week', 'poutcome']¶

The information is written on the basis of EDA and statistical summary¶

Age:-¶

The youngest client is 17 year old and the oldest has 98 years with a median of 38 years and the average is 40 years old. The distribution is skewed. This possibly indicates the some outliers.¶

Marital:-¶

Marital has 4 unique values. these four are:-(married, divorced,single,unknown). Married people has more subscription i;e 60.5%, single has (28.1%), divorced has (11.2%) and unknown has (0.2%), According to piechart and countplot.¶

Education:-¶

Education has 8 unique values. University_degree student has more subscription i;e (29.5%). the four education levels (university_degree, high_school, basic.9y and proffessional_course) correspond 80% of the data. clients with 4 basics or illiterate are the oldest and prone to subscribe to the product.¶

Default:-¶

The client has no credits have more subscription, the subscription percentage has 79.1% a/c to pie and countplot.¶

Housing:-¶

This value has 3 unique value with 'unknown' interpretd as a missing value, representing 2% of observation¶

Job:-¶

The 5 most common jobs are enough to represent 80% of the data.The classes ‘admin.’, ‘blue-collar’, ‘technician’, ‘management’ and ‘ services’ are the jobs where most clients had to subscribe but in terms of proportion, ‘student’ and ‘retired’ are the most representative. we see that ‘retired’ and ‘housemaid’ are the oldest clients and the ones who have accepted the subscription more than any of the other classes. its also so that some unknown values.it mean that they are not mention job type.¶

Loan:-¶

Loan show high number of non-subscribers.¶

Contact:-¶

Acoording to graph the client has more contacted with cellular as compared to other. cellular has 63.5% and telephone has 36.5%¶

Month:-¶

May has more number as compaqared to other 12 months. may has 33.4%¶

Day_of_week:-¶

Day_of_week has 7 aatributes, Thu has little more as compared to Mon. It's distribution show that there is no outliers and no missing values.¶

Duration:-¶

Mostly duration is in seconds. the mean is 258.285 and median is 180. it show that the more difference in between mean and median¶

Campaign:-¶

In distribution graph show that skewed in right. Total 56 campaing has done. skewed in grath it show that outliers presence in campain.¶

Pday:-¶

The majority of client have 999 number which indicates most people did not contact or not conacted by the bank. Those 999 are cosidered to be out of ranges.¶

Previous:-¶

It is about the number of contacts performed before this campaign and for this client¶

Poutcome:-¶

Non-existance customer has no subscription is more number.¶

Emp_var_rate:-¶

employment variation rate. During the period index varied from -3.4 to 1.4¶

Cons_price_index:-¶

It is about the consumer price index for monthly period. The mean is 93.5757 and median is 93.749. there is only small difference between mean and the median.¶

Cons_conf_index:-¶

the consumer confidence level during that period kept always negative with a range of variation of [-51, -27]. These negative values might be explained by the recession that severely affected Portugal due to the financial global crisis during that same period the data was recorded.¶

Euribor3m:-¶

mean is 3.6213 and median is 4.857.¶

nr.employed:-¶

The number of employed people varied around 200 during the campaign.¶

y:-¶

y represent outcome of data means target of the data. the term deposit represent the client are subscribe has represent in ('yes' or 'no'. In EDA graph , highest count is no and lowest count is yes.¶

DATA PREPROCESSING AND FEATURE ENGINEERING¶

In [33]:
# Checking correlation

plt.figure(figsize=(10,20))
sns.heatmap(data.corr(), annot=True, linewidth=0.03)
Out[33]:
<Axes: >
In [34]:
data.corr()
Out[34]:
age duration campaign pdays previous emp_var_rate cons_price_idx cons_conf_idx euribor3m nr_employed y
age 1.000000 -0.000866 0.004594 -0.034369 0.024365 -0.000371 0.000857 0.129372 0.010767 -0.017725 0.030399
duration -0.000866 1.000000 -0.071699 -0.047577 0.020640 -0.027968 0.005312 -0.008173 -0.032897 -0.044703 0.405274
campaign 0.004594 -0.071699 1.000000 0.052584 -0.079141 0.150754 0.127836 -0.013733 0.135133 0.144095 -0.066357
pdays -0.034369 -0.047577 0.052584 1.000000 -0.587514 0.271004 0.078889 -0.091342 0.296899 0.372605 -0.324914
previous 0.024365 0.020640 -0.079141 -0.587514 1.000000 -0.420489 -0.203130 -0.050936 -0.454494 -0.501333 0.230181
emp_var_rate -0.000371 -0.027968 0.150754 0.271004 -0.420489 1.000000 0.775334 0.196041 0.972245 0.906970 -0.298334
cons_price_idx 0.000857 0.005312 0.127836 0.078889 -0.203130 0.775334 1.000000 0.058986 0.688230 0.522034 -0.136211
cons_conf_idx 0.129372 -0.008173 -0.013733 -0.091342 -0.050936 0.196041 0.058986 1.000000 0.277686 0.100513 0.054878
euribor3m 0.010767 -0.032897 0.135133 0.296899 -0.454494 0.972245 0.688230 0.277686 1.000000 0.945154 -0.307771
nr_employed -0.017725 -0.044703 0.144095 0.372605 -0.501333 0.906970 0.522034 0.100513 0.945154 1.000000 -0.354678
y 0.030399 0.405274 -0.066357 -0.324914 0.230181 -0.298334 -0.136211 0.054878 -0.307771 -0.354678 1.000000

Checking null values¶

In [35]:
data.isnull().sum()
Out[35]:
age               0
job               0
marital           0
education         0
default           0
housing           0
loan              0
contact           0
month             0
day_of_week       0
duration          0
campaign          0
pdays             0
previous          0
poutcome          0
emp_var_rate      0
cons_price_idx    0
cons_conf_idx     0
euribor3m         0
nr_employed       0
y                 0
dtype: int64

Data has no null values¶

Checking outliers¶

In [36]:
# Checking outliers in ['Age']
sns.boxplot(data['age'])
Out[36]:
<Axes: >
In [37]:
Q1 = data['age'].quantile(q=.25)
Q3 = data['age'].quantile(q=.75)
print('Q1 is: ', Q1)
print('Q3 is: ', Q3)

Q1 is:  32.0
Q3 is:  47.0
In [38]:
# IQR = Q3-Q1
# Lower 1.5*IQR  IS Q1-1.5*IQR
# Upper 1.5*IQR  IS Q3+1.5*IQR
l_outlier = Q1-1.5*(Q3-Q1)
u_outlier = Q3+1.5*(Q3-Q1)
print('l_outlier is: ', l_outlier)
print('u_outlier is: ', u_outlier)

l_outlier is:  9.5
u_outlier is:  69.5
In [39]:
# number of outliers

print('number of outliers in age upper: ', data[data['age']>69.5]['age'].count())
print('number of outliers in age lower: ',data[data['age']<9.5]['age'].count())

number of outliers in age upper:  469
number of outliers in age lower:  0
In [40]:
data.loc[data['age']>69.5]
Out[40]:
age job marital education default housing loan contact month day_of_week ... campaign pdays previous poutcome emp_var_rate cons_price_idx cons_conf_idx euribor3m nr_employed y
27713 70 retired divorced basic.4y no yes no cellular mar mon ... 3 999 0 nonexistent -1.8 92.843 -50.0 1.811 5099.1 1
27757 76 retired married university.degree no no yes cellular mar thu ... 9 999 0 nonexistent -1.8 92.843 -50.0 1.757 5099.1 0
27780 73 retired married university.degree no yes no cellular mar tue ... 1 999 1 failure -1.8 92.843 -50.0 1.687 5099.1 0
27800 88 retired divorced basic.4y no yes no cellular mar wed ... 1 999 0 nonexistent -1.8 92.843 -50.0 1.663 5099.1 0
27802 88 retired divorced basic.4y no no no cellular mar wed ... 2 999 0 nonexistent -1.8 92.843 -50.0 1.663 5099.1 1
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
40986 84 retired divorced basic.4y unknown yes yes cellular oct mon ... 4 3 1 success -1.1 94.601 -49.5 1.000 4963.6 0
40996 81 retired married basic.4y no yes no cellular oct wed ... 1 999 2 failure -1.1 94.601 -49.5 1.016 4963.6 1
41004 80 retired married professional.course no yes no cellular oct thu ... 1 999 1 failure -1.1 94.601 -49.5 1.025 4963.6 1
41183 73 retired married professional.course no yes no cellular nov fri ... 1 999 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 1
41187 74 retired married professional.course no yes no cellular nov fri ... 3 999 1 failure -1.1 94.767 -50.8 1.028 4963.6 0

469 rows × 21 columns

In [41]:
median = data.loc[data['age']<69.5, 'age'].median()
median
Out[41]:
38.0
In [42]:
data.loc[data.age>69.5, 'age']=np.nan
data['age'].fillna(data['age'].median(), inplace=True)
In [43]:
sns.boxplot(data['age'])
Out[43]:
<Axes: >
In [44]:
# checking outlier in duration
sns.boxplot(data['duration'])
Out[44]:
<Axes: >
In [45]:
# duration

Q1 = data['duration'].quantile(q=.25)
Q3 = data['duration'].quantile(q=.75)
print('Q1 is: ', Q1)
print('Q3 is: ', Q3)

Q1 is:  102.0
Q3 is:  319.0
In [46]:
# IQR = Q3-Q1
# LOWER = 1.5*IQR IS Q1-1.5*IQR
# UPPER = 1.5*IQR IS Q3+1.5*IQR
l_outlier = Q1-1.5*(Q3-Q1)
u_outlier = Q3+1.5*(Q3-Q1)
print('l_outlier is: ', l_outlier)
print('u_outlier is: ', u_outlier)

l_outlier is:  -223.5
u_outlier is:  644.5
In [47]:
print('Number of outlier in duration upper is: ', data[data['duration']>644.5]['duration'].count())
print('number of outlier in duration lower is: ', data[data['duration']<-223.5]['duration'].count())

Number of outlier in duration upper is:  2963
number of outlier in duration lower is:  0
In [48]:
data.loc[data['duration']>644.5]
Out[48]:
age job marital education default housing loan contact month day_of_week ... campaign pdays previous poutcome emp_var_rate cons_price_idx cons_conf_idx euribor3m nr_employed y
37 52.0 technician married basic.9y no yes no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
57 45.0 services married high.school unknown yes no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
61 51.0 blue-collar married basic.9y no yes no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
75 41.0 blue-collar divorced basic.4y unknown yes no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 1
83 49.0 entrepreneur married university.degree unknown yes no telephone may mon ... 1 999 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 1
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
41147 25.0 technician single professional.course no yes no cellular nov fri ... 2 19 1 success -1.1 94.767 -50.8 1.040 4963.6 1
41153 67.0 housemaid divorced professional.course no yes no cellular nov mon ... 2 5 5 success -1.1 94.767 -50.8 1.039 4963.6 1
41160 33.0 admin. married university.degree no no no cellular nov tue ... 1 999 0 nonexistent -1.1 94.767 -50.8 1.035 4963.6 1
41164 54.0 admin. married professional.course no no no cellular nov tue ... 2 10 1 success -1.1 94.767 -50.8 1.035 4963.6 1
41166 32.0 admin. married university.degree no no no telephone nov wed ... 1 999 1 failure -1.1 94.767 -50.8 1.030 4963.6 1

2963 rows × 21 columns

In [49]:
median = data.loc[data['duration']<644.5, 'duration'].median()
median
Out[49]:
167.0
In [50]:
data.loc[data['duration']>644.5, 'duration']=np.nan
data['duration'].fillna(data['duration'].median(), inplace = True)
In [51]:
sns.boxplot(data['duration'])
Out[51]:
<Axes: >
In [52]:
# Campaign 
sns.boxplot(data['campaign'])
Out[52]:
<Axes: >
In [53]:
Q1 = data['campaign'].quantile(q=.25)
Q3 = data['campaign'].quantile(q=.75)
print('Q1 is: ', Q1)
print('Q3 is: ', Q3)

Q1 is:  1.0
Q3 is:  3.0
In [54]:
# IQR =Q3-Q1
# LOWER = 1.5*IQR IS Q1-1.5*IQR
# UPPER = 1.5*IQR IS Q3-1.5*IQR
l_outlier = Q1- 1.5*(Q3-Q1)
u_outlier = Q3+ 1.5*(Q3-Q1)
print('l_outlier is: ', l_outlier)
print('u_outlier is: ', u_outlier)

l_outlier is:  -2.0
u_outlier is:  6.0
In [55]:
print('number of outlier in campaign upper is: ', data[data['campaign']>6.0]['campaign'].count())
print('number of outlier in camapaign lower is: ', data[data['campaign']<(-2.0)]['campaign'].count())

number of outlier in campaign upper is:  2406
number of outlier in camapaign lower is:  0
In [56]:
data.loc[data['campaign']>6.0]
Out[56]:
age job marital education default housing loan contact month day_of_week ... campaign pdays previous poutcome emp_var_rate cons_price_idx cons_conf_idx euribor3m nr_employed y
887 54.0 admin. married university.degree no no no telephone may wed ... 7 999 0 nonexistent 1.1 93.994 -36.4 4.856 5191.0 0
1043 41.0 technician single university.degree unknown no no telephone may wed ... 8 999 0 nonexistent 1.1 93.994 -36.4 4.856 5191.0 0
1094 40.0 technician married professional.course no yes no telephone may wed ... 8 999 0 nonexistent 1.1 93.994 -36.4 4.856 5191.0 0
1097 51.0 blue-collar married basic.4y unknown yes no telephone may wed ... 7 999 0 nonexistent 1.1 93.994 -36.4 4.856 5191.0 0
1230 48.0 blue-collar divorced basic.4y no no no telephone may thu ... 7 999 0 nonexistent 1.1 93.994 -36.4 4.855 5191.0 0
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
40611 38.0 retired married basic.9y no no no telephone sep fri ... 9 999 1 failure -1.1 94.199 -37.5 0.883 4963.6 0
40631 38.0 retired married basic.4y no yes no cellular sep mon ... 7 999 0 nonexistent -1.1 94.199 -37.5 0.882 4963.6 0
40698 29.0 technician single professional.course no yes no cellular sep mon ... 7 3 5 success -1.1 94.199 -37.5 0.879 4963.6 0
40821 33.0 technician single professional.course no yes no cellular sep fri ... 9 999 2 failure -1.1 94.199 -37.5 0.879 4963.6 0
40993 50.0 entrepreneur divorced university.degree no yes no telephone oct tue ... 7 11 2 success -1.1 94.601 -49.5 1.008 4963.6 1

2406 rows × 21 columns

In [57]:
median = data.loc[data['campaign']<6.0, 'campaign'].median()
median
Out[57]:
2.0
In [58]:
data.loc[data['campaign']>6.0, 'campaign']=np.nan
data['campaign'].fillna(data['campaign'].median(), inplace=True)
In [59]:
sns.boxplot(data['campaign'])
Out[59]:
<Axes: >
In [60]:
# Checking pdays
sns.boxplot(data['pdays'])
Out[60]:
<Axes: >
In [61]:
Q1 = data['pdays'].quantile(q=.25)
Q3 = data['pdays'].quantile(q=.75)
print('Q1 is: ', Q1)
print('Q3 is: ', Q3)

Q1 is:  999.0
Q3 is:  999.0
In [62]:
# IQR = Q3-Q1
# upper = 1.5*IQR IS Q3+1.5*IQR
# lower = 1.5*IQR IS Q1-1.5*IQR
l_outlier = Q3+1.5*(Q3-Q1)
u_outlier = Q1- 1.5*(Q3-Q1)
print('l_outlier is: ', l_outlier)
print('u_outlier is: ', u_outlier)

l_outlier is:  999.0
u_outlier is:  999.0
In [63]:
print('number of outlier in pdays upper is: ', data[data['pdays']>999.0]['pdays'].count())
print('number of outlier in pdays lower is: ', data[data['pdays']<999.0]['pdays'].count())

number of outlier in pdays upper is:  0
number of outlier in pdays lower is:  1515
In [64]:
data.loc[data['pdays']<999.0]
Out[64]:
age job marital education default housing loan contact month day_of_week ... campaign pdays previous poutcome emp_var_rate cons_price_idx cons_conf_idx euribor3m nr_employed y
24108 37.0 admin. married university.degree no yes no telephone nov wed ... 1.0 6 1 success -0.1 93.200 -42.0 4.286 5195.8 0
24264 48.0 management married university.degree no yes no cellular nov mon ... 1.0 4 1 success -0.1 93.200 -42.0 4.191 5195.8 0
24279 42.0 technician married professional.course unknown yes no cellular nov mon ... 1.0 4 1 success -0.1 93.200 -42.0 4.191 5195.8 0
24397 50.0 entrepreneur married university.degree unknown yes no cellular nov mon ... 1.0 3 1 success -0.1 93.200 -42.0 4.191 5195.8 0
24482 36.0 self-employed single university.degree no yes no cellular nov mon ... 2.0 4 1 success -0.1 93.200 -42.0 4.191 5195.8 0
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
41163 35.0 technician divorced basic.4y no yes no cellular nov tue ... 3.0 4 2 success -1.1 94.767 -50.8 1.035 4963.6 1
41164 54.0 admin. married professional.course no no no cellular nov tue ... 2.0 10 1 success -1.1 94.767 -50.8 1.035 4963.6 1
41174 62.0 retired married university.degree no yes no cellular nov thu ... 1.0 1 6 success -1.1 94.767 -50.8 1.031 4963.6 1
41178 62.0 retired married university.degree no no no cellular nov thu ... 2.0 6 3 success -1.1 94.767 -50.8 1.031 4963.6 1
41182 29.0 unemployed single basic.4y no yes no cellular nov fri ... 1.0 9 1 success -1.1 94.767 -50.8 1.028 4963.6 0

1515 rows × 21 columns

In [65]:
median = data.loc[data['pdays']<999.0, 'pdays'].median()
median
Out[65]:
6.0
In [66]:
data.loc[data['pdays']<999.0, 'pdays']=np.nan
data['pdays'].fillna(data['pdays'].median(), inplace=True)
In [67]:
sns.boxplot(data['pdays'])
Out[67]:
<Axes: >
In [68]:
# Checking outlier in cons_conf_idx
sns.boxplot(data['cons_conf_idx'])
Out[68]:
<Axes: >
In [69]:
Q1 = data['cons_conf_idx'].quantile(.25)
Q3 = data['cons_conf_idx'].quantile(.75)
print('Q1 is: ', Q1)
print('Q3 is: ', Q3)

Q1 is:  -42.7
Q3 is:  -36.4
In [70]:
# IQR = Q3-Q1
# lower = 1.5*IQR is Q1-1.5*IQR
# upper = 1.5*IQR is Q3-1.5*IQR
l_outlier = Q1-1.5*(Q3-Q1)
u_outlier = Q3+1.5*(Q3-Q1)
print('l_outlier is: ', l_outlier)
print('u_outlier is: ', u_outlier)

l_outlier is:  -52.150000000000006
u_outlier is:  -26.949999999999992
In [71]:
print('number of lower outlier in cons_conf_idx is: ', data[data['cons_conf_idx']<(-52.150000000000006)]['cons_conf_idx'].count())
print('number of upper outlier in cons_conf_idx is: ', data[data['cons_conf_idx']>(-26.949999999999992)]['cons_conf_idx'].count())
      

number of lower outlier in cons_conf_idx is:  0
number of upper outlier in cons_conf_idx is:  447
In [72]:
data.loc[data['cons_conf_idx']> (-26.949999999999992)]
Out[72]:
age job marital education default housing loan contact month day_of_week ... campaign pdays previous poutcome emp_var_rate cons_price_idx cons_conf_idx euribor3m nr_employed y
38154 50.0 management married university.degree no yes no cellular oct thu ... 2.0 999.0 1 success -3.4 92.431 -26.9 0.754 5017.5 1
38155 37.0 admin. single university.degree no yes no cellular oct thu ... 3.0 999.0 0 nonexistent -3.4 92.431 -26.9 0.754 5017.5 1
38156 59.0 technician single basic.6y no no no cellular oct thu ... 1.0 999.0 2 failure -3.4 92.431 -26.9 0.754 5017.5 0
38157 31.0 admin. married university.degree no yes no cellular oct thu ... 1.0 999.0 0 nonexistent -3.4 92.431 -26.9 0.754 5017.5 0
38158 35.0 admin. married high.school no yes no cellular oct thu ... 1.0 999.0 1 success -3.4 92.431 -26.9 0.754 5017.5 1
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
38596 69.0 retired married basic.4y no yes yes cellular oct fri ... 3.0 999.0 0 nonexistent -3.4 92.431 -26.9 0.720 5017.5 1
38597 18.0 student single basic.6y no no yes cellular oct fri ... 2.0 999.0 0 nonexistent -3.4 92.431 -26.9 0.720 5017.5 1
38598 59.0 retired divorced basic.4y no yes no telephone oct fri ... 4.0 999.0 0 nonexistent -3.4 92.431 -26.9 0.720 5017.5 0
38599 37.0 admin. single university.degree no no no cellular oct fri ... 2.0 999.0 0 nonexistent -3.4 92.431 -26.9 0.720 5017.5 1
38600 38.0 retired divorced basic.6y no no no telephone oct fri ... 2.0 999.0 0 nonexistent -3.4 92.431 -26.9 0.720 5017.5 0

447 rows × 21 columns

In [73]:
median= data.loc[data['cons_conf_idx']<(-26.949999999999992),'cons_conf_idx'].median()
median
Out[73]:
-41.8
In [74]:
data.loc[data['cons_conf_idx']>(-26.949999999999992), 'cons_conf_idx']=np.nan
data['cons_conf_idx'].fillna(data['cons_conf_idx'].median(), inplace=True)
In [75]:
sns.boxplot(data['cons_conf_idx'])
Out[75]:
<Axes: >

Converting all categorical data to numerical¶

In [76]:
data
Out[76]:
age job marital education default housing loan contact month day_of_week ... campaign pdays previous poutcome emp_var_rate cons_price_idx cons_conf_idx euribor3m nr_employed y
0 56.0 housemaid married basic.4y no no no telephone may mon ... 1.0 999.0 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
1 57.0 services married high.school unknown no no telephone may mon ... 1.0 999.0 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
2 37.0 services married high.school no yes no telephone may mon ... 1.0 999.0 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
3 40.0 admin. married basic.6y no no no telephone may mon ... 1.0 999.0 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
4 56.0 services married high.school no no yes telephone may mon ... 1.0 999.0 0 nonexistent 1.1 93.994 -36.4 4.857 5191.0 0
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
41183 38.0 retired married professional.course no yes no cellular nov fri ... 1.0 999.0 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 1
41184 46.0 blue-collar married professional.course no no no cellular nov fri ... 1.0 999.0 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 0
41185 56.0 retired married university.degree no yes no cellular nov fri ... 2.0 999.0 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 0
41186 44.0 technician married professional.course no no no cellular nov fri ... 1.0 999.0 0 nonexistent -1.1 94.767 -50.8 1.028 4963.6 1
41187 38.0 retired married professional.course no yes no cellular nov fri ... 3.0 999.0 1 failure -1.1 94.767 -50.8 1.028 4963.6 0

41188 rows × 21 columns

In [77]:
df=data.copy()
In [78]:
df['job'].replace(['housemaid','services','admin.', 'blue-collar', 'technician' ,'retired','management', 'unemployed', 'self-employed', 'unknown' ,'entrepreneur','student'],[0,1,2,3,4,5,6,7,8,9,10,11],inplace=True)
df['default'].replace([ 'no','yes','unknown'],[7,8,9], inplace=True)
df['education'].replace(['basic.4y','basic.6y','basic.9y','professional.course', 'unknown', 'high.school','university.degree','illiterate'],[3,2,0,1,4,5,6,7],inplace=True)
df['marital'].replace([ 'divorced','married','single','unknown'],[10,11,12,13], inplace=True)
df['housing'].replace(['no','yes','unknown'],[15,16,17], inplace=True)
df['loan'].replace([ 'no','yes','unknown'],[0,2,4], inplace=True)
df['contact'].replace([ 'cellular','telephone'], [7,9], inplace=True)
df['month'].replace(['may', 'jun', 'jul', 'aug', 'oct', 'nov', 'dec', 'jan', 'feb', 'mar', 'apr', 'sep'],[5,6,7,8,10,11,12,1,2,3,4,9],inplace=True)
df['day_of_week'].replace([ 'mon','tue','wed','thu','fri'], [7,6,5,4,3], inplace=True)
df['poutcome'].replace(['failure','nonexistent','success'],[4,2,7], inplace=True)
In [79]:
df
Out[79]:
age job marital education default housing loan contact month day_of_week ... campaign pdays previous poutcome emp_var_rate cons_price_idx cons_conf_idx euribor3m nr_employed y
0 56.0 0 11 3 7 15 0 9 5 7 ... 1.0 999.0 0 2 1.1 93.994 -36.4 4.857 5191.0 0
1 57.0 1 11 5 9 15 0 9 5 7 ... 1.0 999.0 0 2 1.1 93.994 -36.4 4.857 5191.0 0
2 37.0 1 11 5 7 16 0 9 5 7 ... 1.0 999.0 0 2 1.1 93.994 -36.4 4.857 5191.0 0
3 40.0 2 11 2 7 15 0 9 5 7 ... 1.0 999.0 0 2 1.1 93.994 -36.4 4.857 5191.0 0
4 56.0 1 11 5 7 15 2 9 5 7 ... 1.0 999.0 0 2 1.1 93.994 -36.4 4.857 5191.0 0
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
41183 38.0 5 11 1 7 16 0 7 11 3 ... 1.0 999.0 0 2 -1.1 94.767 -50.8 1.028 4963.6 1
41184 46.0 3 11 1 7 15 0 7 11 3 ... 1.0 999.0 0 2 -1.1 94.767 -50.8 1.028 4963.6 0
41185 56.0 5 11 6 7 16 0 7 11 3 ... 2.0 999.0 0 2 -1.1 94.767 -50.8 1.028 4963.6 0
41186 44.0 4 11 1 7 15 0 7 11 3 ... 1.0 999.0 0 2 -1.1 94.767 -50.8 1.028 4963.6 1
41187 38.0 5 11 1 7 16 0 7 11 3 ... 3.0 999.0 1 4 -1.1 94.767 -50.8 1.028 4963.6 0

41188 rows × 21 columns

In [80]:
X=df.drop(columns=['y'])
Y=df['y']
In [81]:
X
Out[81]:
age job marital education default housing loan contact month day_of_week duration campaign pdays previous poutcome emp_var_rate cons_price_idx cons_conf_idx euribor3m nr_employed
0 56.0 0 11 3 7 15 0 9 5 7 261.0 1.0 999.0 0 2 1.1 93.994 -36.4 4.857 5191.0
1 57.0 1 11 5 9 15 0 9 5 7 149.0 1.0 999.0 0 2 1.1 93.994 -36.4 4.857 5191.0
2 37.0 1 11 5 7 16 0 9 5 7 226.0 1.0 999.0 0 2 1.1 93.994 -36.4 4.857 5191.0
3 40.0 2 11 2 7 15 0 9 5 7 151.0 1.0 999.0 0 2 1.1 93.994 -36.4 4.857 5191.0
4 56.0 1 11 5 7 15 2 9 5 7 307.0 1.0 999.0 0 2 1.1 93.994 -36.4 4.857 5191.0
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
41183 38.0 5 11 1 7 16 0 7 11 3 334.0 1.0 999.0 0 2 -1.1 94.767 -50.8 1.028 4963.6
41184 46.0 3 11 1 7 15 0 7 11 3 383.0 1.0 999.0 0 2 -1.1 94.767 -50.8 1.028 4963.6
41185 56.0 5 11 6 7 16 0 7 11 3 189.0 2.0 999.0 0 2 -1.1 94.767 -50.8 1.028 4963.6
41186 44.0 4 11 1 7 15 0 7 11 3 442.0 1.0 999.0 0 2 -1.1 94.767 -50.8 1.028 4963.6
41187 38.0 5 11 1 7 16 0 7 11 3 239.0 3.0 999.0 1 4 -1.1 94.767 -50.8 1.028 4963.6

41188 rows × 20 columns

Scaling the data¶

In [82]:
# Scaling the data as all feature seems to be near to normal distribution

from sklearn.preprocessing import StandardScaler
Scaler= StandardScaler() # object creation
X_scaled = Scaler.fit_transform(X)
In [83]:
X_scaled
Out[83]:
array([[ 1.70658113, -1.49696895, -0.2837415 , ...,  0.9680674 ,
         0.71245988,  0.33167991],
       [ 1.81051927, -1.08948536, -0.2837415 , ...,  0.9680674 ,
         0.71245988,  0.33167991],
       [-0.26824358, -1.08948536, -0.2837415 , ...,  0.9680674 ,
         0.71245988,  0.33167991],
       ...,
       [ 1.70658113,  0.540449  , -0.2837415 , ..., -2.30096943,
        -1.49518647, -2.8156966 ],
       [ 0.45932342,  0.13296541, -0.2837415 , ..., -2.30096943,
        -1.49518647, -2.8156966 ],
       [-0.16430544,  0.540449  , -0.2837415 , ..., -2.30096943,
        -1.49518647, -2.8156966 ]])
In [84]:
from sklearn.model_selection import train_test_split
X_train, X_test ,Y_train, Y_test =train_test_split(X_scaled,Y, test_size=0.25, random_state=42 )

MODEL CREATION¶

LOGISTIC REGRESSION¶

In [85]:
from sklearn.linear_model import LogisticRegression
log_reg = LogisticRegression()

log_reg.fit(X_train,Y_train)
Out[85]:
LogisticRegression()
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LogisticRegression()
In [86]:
# Predicting Y_train
Y_train_pred = log_reg.predict(X_train) 
Y_train_pred
Out[86]:
array([0, 0, 0, ..., 1, 0, 0], dtype=int64)
In [87]:
# Testing model
Y_pred = log_reg.predict(X_test)
In [88]:
Y_pred
Out[88]:
array([0, 0, 0, ..., 0, 0, 0], dtype=int64)
In [89]:
# Known the shape 
Y_train.shape
Out[89]:
(30891,)
In [90]:
Y_pred.shape
Out[90]:
(10297,)

CALCULATING ACCURACY¶

In [91]:
from sklearn.metrics import accuracy_score,confusion_matrix,precision_score,recall_score,f1_score,classification_report
accuracy = accuracy_score(Y_train, Y_train_pred)
accuracy
Out[91]:
0.9033051697905539

In logistic regression algorithm got 90% accuracy¶

KNN¶

In [92]:
from imblearn.over_sampling import SMOTE
smote=SMOTE() # Object creatuin
In [93]:
X_train_smote, Y_train_smote = smote.fit_resample(X_train.astype('float'),Y_train)
In [94]:
from collections import Counter
print('Actual classes', Counter(Y_train))
print('Actual classes', Counter(Y_train_smote))

Actual classes Counter({0: 27404, 1: 3487})
Actual classes Counter({0: 27404, 1: 27404})
In [95]:
from sklearn.neighbors import KNeighborsClassifier

knn = KNeighborsClassifier(n_neighbors=7)
knn.fit(X_train_smote, Y_train_smote)
Out[95]:
KNeighborsClassifier(n_neighbors=7)
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KNeighborsClassifier(n_neighbors=7)
In [96]:
Y_pred = knn.predict(X_test)
In [97]:
error_rate = []
for i in range(1,11):
    knn = KNeighborsClassifier(n_neighbors=i)
    knn.fit(X_train_smote,Y_train_smote)
    pred_i = knn.predict(X_test)
    error_rate.append(np.mean(pred_i != Y_test))
In [98]:
plt.figure(figsize=(10,6))
plt.plot(range(1,11),error_rate,color='green',linestyle='dashed',marker='s',markerfacecolor='purple',markersize=10)
plt.title('Error Rate vs. K value')
plt.xlabel('k')
plt.ylabel('Error Rate')
Out[98]:
Text(0, 0.5, 'Error Rate')
In [99]:
knn=KNeighborsClassifier(n_neighbors=5)
knn.fit(X_train_smote,Y_train_smote)
Y_pred=knn.predict(X_test)
In [100]:
print("The accuracy score is : ", accuracy_score(Y_test,Y_pred))

The accuracy score is :  0.7686704865494804

Knn got 76% accuracy¶

SVM¶

In [101]:
from sklearn.svm import SVC
svm=SVC()
svm.fit(X_train,Y_train)
Out[101]:
SVC()
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SVC()
In [102]:
Y_pred_sv=svm.predict(X_test)
In [103]:
print(accuracy_score(Y_test,Y_pred_sv))

0.9052151111974361

SVM got 91% performance¶

RANDOM FOREST¶

In [104]:
from sklearn.ensemble import RandomForestClassifier
R_forest = RandomForestClassifier()
R_forest.fit(X_train,Y_train)
Out[104]:
RandomForestClassifier()
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RandomForestClassifier()
In [105]:
Y_pred = R_forest.predict(X_test)
Y_pred
Out[105]:
array([0, 0, 0, ..., 0, 0, 0], dtype=int64)
In [106]:
(Y_test == Y_pred).sum()/len(Y_test)*100
Out[106]:
91.19160920656502
In [107]:
print('accuracy_score is: ', accuracy_score(Y_test,Y_pred))

accuracy_score is:  0.9119160920656502

Random forest got 91%¶

XG BOOSTER¶

In [108]:
from sklearn.ensemble import GradientBoostingClassifier
XGB = GradientBoostingClassifier()
XGB.fit(X_train,Y_train)
Out[108]:
GradientBoostingClassifier()
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GradientBoostingClassifier()
In [109]:
Y_pred = XGB.predict(X_test)
Y_pred
Out[109]:
array([0, 0, 0, ..., 0, 0, 0], dtype=int64)
In [110]:
print('Accuracy_score is: ', accuracy_score(Y_test, Y_pred))

Accuracy_score is:  0.9119160920656502

XGB is got 92% performance¶

DECISION TREE¶

In [111]:
from sklearn.tree import DecisionTreeClassifier
d_tree = DecisionTreeClassifier()
d_tree.fit(X_train,Y_train)
Out[111]:
DecisionTreeClassifier()
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DecisionTreeClassifier()
In [112]:
Y_pred = d_tree.predict(X_test)
Y_pred
Out[112]:
array([0, 0, 0, ..., 0, 0, 0], dtype=int64)
In [113]:
accuracy = accuracy_score(Y_test,Y_pred)
accuracy
Out[113]:
0.8864717878993882
In [114]:
from sklearn.metrics import classification_report,accuracy_score,f1_score,recall_score
In [115]:
CR = classification_report(Y_test,Y_pred)
CR
Out[115]:
'              precision    recall  f1-score   support\n\n           0       0.94      0.94      0.94      9144\n           1       0.49      0.50      0.50      1153\n\n    accuracy                           0.89     10297\n   macro avg       0.72      0.72      0.72     10297\nweighted avg       0.89      0.89      0.89     10297\n'
In [116]:
F1_score = f1_score(Y_test,Y_pred)
In [117]:
F1_score
Out[117]:
0.4972043010752688

Decision tree got 88% accuracy¶

CONCLUSION¶

In Machine Learning Project,First:- I have focused on one of the most important part that is EDA(Exploratory data analysis). This is very important procedure to understand the data.¶

Second Focus on Feature engineering and feature selection, cleaning the dataset and taking care of anomalies. we find along the way as well as imputations and transformation¶

Third we use different machine learning algorithm to predict the dataset. like:- Logistic regression, SVM, KNN, Decision Tree, Random Forest and XG Booster.¶

We got maximum in XG BOOSTER, GOT 92%.¶

In [118]:
pwd
Out[118]:
'C:\\Users\\amanp\\PORTUGUESE_BANK_PROJECT_1'
In [119]:
import plotly.express as px

# for html export
import plotly.io as pio
pio.renderers.default='notebook'

# for pdf export
!pip install Pyppeteer
!pyppeteer-install

Collecting Pyppeteer
  Downloading pyppeteer-1.0.2-py3-none-any.whl (83 kB)
     ---------------------------------------- 83.4/83.4 kB 4.9 MB/s eta 0:00:00
Requirement already satisfied: importlib-metadata>=1.4 in c:\users\amanp\anaconda3\lib\site-packages (from Pyppeteer) (4.11.3)
Requirement already satisfied: urllib3<2.0.0,>=1.25.8 in c:\users\amanp\anaconda3\lib\site-packages (from Pyppeteer) (1.26.14)
Collecting websockets<11.0,>=10.0
  Downloading websockets-10.4-cp310-cp310-win_amd64.whl (101 kB)
     ---------------------------------------- 101.4/101.4 kB ? eta 0:00:00
Requirement already satisfied: tqdm<5.0.0,>=4.42.1 in c:\users\amanp\anaconda3\lib\site-packages (from Pyppeteer) (4.64.1)
Collecting pyee<9.0.0,>=8.1.0
  Downloading pyee-8.2.2-py2.py3-none-any.whl (12 kB)
Requirement already satisfied: certifi>=2021 in c:\users\amanp\anaconda3\lib\site-packages (from Pyppeteer) (2022.12.7)
Requirement already satisfied: appdirs<2.0.0,>=1.4.3 in c:\users\amanp\anaconda3\lib\site-packages (from Pyppeteer) (1.4.4)
Requirement already satisfied: zipp>=0.5 in c:\users\amanp\anaconda3\lib\site-packages (from importlib-metadata>=1.4->Pyppeteer) (3.11.0)
Requirement already satisfied: colorama in c:\users\amanp\anaconda3\lib\site-packages (from tqdm<5.0.0,>=4.42.1->Pyppeteer) (0.4.6)
Installing collected packages: pyee, websockets, Pyppeteer
Successfully installed Pyppeteer-1.0.2 pyee-8.2.2 websockets-10.4

[INFO] Starting Chromium download.

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 79%|#######9  | 108M/137M [00:11<00:02, 9.53Mb/s]
 80%|########  | 110M/137M [00:11<00:02, 9.84Mb/s]
 81%|########  | 111M/137M [00:11<00:02, 10.2Mb/s]
 82%|########1 | 112M/137M [00:11<00:02, 10.4Mb/s]
 82%|########2 | 113M/137M [00:11<00:02, 9.15Mb/s]
 84%|########3 | 115M/137M [00:11<00:02, 11.1Mb/s]
 85%|########4 | 116M/137M [00:12<00:01, 10.9Mb/s]
 85%|########5 | 117M/137M [00:12<00:01, 10.9Mb/s]
 86%|########6 | 118M/137M [00:12<00:01, 10.9Mb/s]
 87%|########6 | 119M/137M [00:12<00:01, 10.9Mb/s]
 88%|########7 | 120M/137M [00:12<00:01, 10.9Mb/s]
 89%|########8 | 121M/137M [00:12<00:01, 10.9Mb/s]
 89%|########9 | 122M/137M [00:12<00:01, 10.9Mb/s]
 90%|######### | 124M/137M [00:12<00:01, 10.9Mb/s]
 91%|#########1| 125M/137M [00:12<00:01, 10.9Mb/s]
 92%|#########1| 126M/137M [00:13<00:01, 10.9Mb/s]
 93%|#########2| 127M/137M [00:13<00:01, 8.68Mb/s]
 94%|#########4| 129M/137M [00:13<00:00, 11.6Mb/s]
 95%|#########5| 130M/137M [00:13<00:00, 11.4Mb/s]
 96%|#########5| 131M/137M [00:13<00:00, 11.3Mb/s]
 97%|#########6| 133M/137M [00:13<00:00, 11.2Mb/s]
 98%|#########7| 134M/137M [00:13<00:00, 11.1Mb/s]
 98%|#########8| 135M/137M [00:13<00:00, 11.0Mb/s]
 99%|#########9| 136M/137M [00:13<00:00, 11.0Mb/s]
100%|##########| 137M/137M [00:14<00:00, 9.75Mb/s]
[INFO] Beginning extraction
[INFO] Chromium extracted to: C:\Users\amanp\AppData\Local\pyppeteer\pyppeteer\local-chromium\588429
In [ ]: